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PHY 121
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A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
V0 = 20cm/sec
`ds = 120cm/sec
A = 9.8m/sec^2
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
`ds = 120cm/sec
V0 = 20cm/sec
A = 980cm/sec^2
Vf = 0cm/sec
vAve = 10cm/sec
`dv = 20cm/sec
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What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
V0=80cm/sec
`ds = 120cm
A = 980cm/sec^2
Vf^2 = v0^2 + 2(a*`ds)
Vf^2 = (80cm/sec)^2 + 2(980cm/sec^2 * 120cm)
Vf^2 = 241600cm^2/sec^2
Vf = +-(491.53cm/sec)
Vf = +491.53cm/sec
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120 cm is in the vertical direction, as in the 980 cm/s^2 acceleration, so the horizontal velocity 80 cm/s will not be in the same calculation as the other two quantities. The initial vertical velocity is 20 cm/s downward, and should have been used here.
Having found the time of fall, then, you will just multiply the horizontal velocity (which, as you say, is constant) by the time of fall.
The time interval is the same for horizontal and vertical motion.
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`ds = vAve *`dt
120cm = (491.53cm/sec+80cm/sec)/2 * `dt
`dt = .42sec
I calculated this, but it doesn’t seem to make sense to me. I think I remember a problem where it said we consider the horizontal motion to be constant with no acceleration and that is should be considered independent of the vertical motion. If that is the case, both v0 and vf should be the same at 80cm/sec and `dt = 1.5sec
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
V0 = 80cm/sec
`dt = .42sec
A = 980cm/sec^2
Vf = 491.53cm/sec
After .42 seconds traveling at 80cm/sec
`ds = 33.6cm
vAve = (491.53cm/sec+80cm/sec)/2 = 285.77cm/sec
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You would not average a vertical velocity with a horizontal velocity.
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`dv = (491.53cm/sec-80cm/sec) = 411.53cm/sec
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No.
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Why does this analysis stop at the instant of impact with the floor?
Until it hits the floor the only resistance encountered would be negligible air resistance. Once it hits the floor, the impact and subsequent friction will change the acceleration, gravity will no longer be acting on the ball and the system will behave differently.
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Gravity is still acting on the ball. However other forces are also acting, which as you say changes the net force and therefore the acceleration.
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Good overall, but you mixed horizontal quantities with vertical quantities in analyzing the projectile. Should be easy to fix, with my notes.
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